New Paper on OHC?

[skierinvermont]

I was looking for the latest OHC data and came across this graphic from a Levitus paper in preparation. Anybody know anything about this paper?

0-700m is also really rising lately too and finally surpassing what was achieved in the 1998 El Nino. That's important because deniers have traditionally tossed it out the window without considering the two independent lines of evidence (the sea level budget and estimates from ARGO).

The earth remains in a large energy imbalance despite the solar minimum. More energy is coming in than is leaving.

[Msalgado]

http://www.agu.org/p...2GL051106.shtml

I believe this is the paper.

Edit - Better link: http://www.agu.org/pubs/crossref/2012/2012GL051106.shtml

and SS page on it:

http://skepticalscience.com/levitus-2012-global-warming-heating-oceans.html

On a side note, it would be nice to have a "clearinghouse" thread with a lot of the most important recent literature regarding AGW. Would make for a reading list.

[WeatherRusty]

http://www.agu.org/p...2GL051106.shtml

I believe this is the paper.

Edit - Better link: http://www.agu.org/p...2GL051106.shtml

and SS page on it:

http://skepticalscie...ing-oceans.html

On a side note, it would be nice to have a "clearinghouse" thread with a lot of the most important recent literature regarding AGW. Would make for a reading list.

Check this out

AGW Observer

I also added the link to the Climate Change Primer thread.

[Msalgado]

Awesome link. Thank you!

[The_Global_Warmer]

We have had some recent discussions about this here. The deniers basically said the data is invalid, contains to many errors and the the solar min operates on a 7-8 year lag so it has nothing to do with the recent big slow down of OHC

[skierinvermont]

We have had some recent discussions about this here. The deniers basically said the data is invalid, contains to many errors and the the solar min operates on a 7-8 year lag so it has nothing to do with the recent big slow down of OHC

I've heard some propose a 7-8 year lag between solar and surface temperatures (which is false the lag is a year or two at most).

A lag between solar and OHC would be truly magical. Solar is supposed to directly effect the energy budget. The hypothesized lag occurs because the heat goes into the oceans first.

It would be as if changes in the solar variables reaching earth had absolutely no detectable physical effect for 7-8 years and then suddenly started causing warming/cooling.

[ben4vols]

The earth remains in a large energy imbalance despite the solar minimum. More energy is coming in than is leaving.

Can I get a source for this statement?

[csnavywx]

Can I get a source for this statement?

He could. Or you could type it in and get it extremely easily yourself. It took me all of 10 seconds. Stop begging for hand-holding. Put dem skills to work. There's an avalanche of work out there related to it.

[ben4vols]

He could. Or you could type it in and get it extremely easily yourself. It took me all of 10 seconds. Stop begging for hand-holding. Put dem skills to work. There's an avalanche of work out there related to it.

10 seconds? Congrats Google king. I was wondering what source he was using (since there is a wide variety of opinions on this). Maybe my question should have been more specific.

[Vergent]

10 seconds? Congrats Google king. I was wondering what source he was using (since there is a wide variety of opinions on this). Maybe my question should have been more specific.

If you want to know where a graph comes from right click on it and choose "inspect element". scroll down to the dark blue highlighted link and left click on the link. You get a new pop-up with the graph and a link below it. click on that link, thats where it came from.

[skierinvermont]

Can I get a source for this statement?

The Levitus study posted in this thread several times already.

(Increasing OHC is a direct measure of the planetary energy imbalance. More energy is entering the earth system than leaving it. Levitus et al. estimated the imbalance to be .27W/m2 over the last 55 years).

[QVectorman]

The trouble with the paper is that the monthly temperature climatologies deeper than 1500 meters have not been calculated/observed, therefore data at the 2000 meter depth doesn't exist. Add in that in 55 years, only 5% of the 1° X 1° gridcells have three observations or more for January-March at the1500 meter level. So data is a little spars to say the least.

[skierinvermont]

The trouble with the paper is that the monthly temperature climatologies deeper than 1500 meters have not been calculated/observed, therefore data at the 2000 meter depth doesn't exist. Add in that in 55 years, only 5% of the 1° X 1° gridcells have three observations or more for January-March at the1500 meter level. So data is a little spars to say the least.

They present error bars calculated based on the sparsity of data. There is no reason I can think of why these error bars would not present a complete picture of the possible error.

[QVectorman]

The sample size is too small to accurately calculate the standard error of the mean. Calculation of the SEM is the standard deviation divided by the square root of N, sample size. So with a small sample size N the SEM can still be calculated although issues develop. Notice the relation that as N increases SEM gets smalls and vis versa as N decreases the SEM increases. So as the sample size decrease, the normal calculation of the SEM progressively underestimates the SEM more and more. As a result, with there only being on average three data points per location in the sample over a 55 year period, the SEM calculation underestimates the actual standard error of the mean by about 12%. So of instead of 95% of the data being within the 95% confidence interval, only about 80% of the data is in the 95% confidence interval.

[skierinvermont]

The sample size is too small to accurately calculate the standard error of the mean. Calculation of the SEM is the standard deviation divided by the square root of N, sample size. So with a small sample size N the SEM can still be calculated although issues develop. Notice the relation that as N increases SEM gets smalls and vis versa as N decreases the SEM increases. So as the sample size decrease, the normal calculation of the SEM progressively underestimates the SEM more and more. As a result, with there only being on average three data points per location in the sample over a 55 year period, the SEM calculation underestimates the actual standard error of the mean by about 12%. So of instead of 95% of the data being within the 95% confidence interval, only about 80% of the data is in the 95% confidence interval.

Never heard that one before... 'the normal calculation of the SEM underestimates the SEM'... I suggest you go back and re-learn some statistics.

The SEM is not underestimated by its proper calculation. Small N's yield the expected ridiculously large SEM. Fortunately in this case, the N is more than sufficient. The SEM calculation is actually quite complicated because there are both spatial and temporal issues. You rightly point out that many 5 degree X 5 degree grid cells are not well sampled over the long run. Fortunately, one does not need to sample every 5 X 5 degree grid cell globally to come up with a good estimate of global temperature change. As long as one has good data from at least 100 or so of the 2500+ 5X5 degree cells, a reasonable estimate can be made with a reasonably small SEM. In this case, many more than 100 cells have good sampling I would presume, given the fairly small SEM.

These are complicated statistical issues that take time and effort to learn.

[QVectorman]

Never heard that one before... 'the normal calculation of the SEM underestimates the SEM'... I suggest you go back and re-learn some statistics.

The SEM is not underestimated by its proper calculation. Small N's yield the expected ridiculously large SEM. Fortunately in this case, the N is more than sufficient. The SEM calculation is actually quite complicated because there are both spatial and temporal issues. You rightly point out that many 5 degree X 5 degree grid cells are not well sampled over the long run. Fortunately, one does not need to sample every 5 X 5 degree grid cell globally to come up with a good estimate of global temperature change. As long as one has good data from at least 100 or so of the 2500+ 5X5 degree cells, a reasonable estimate can be made with a reasonably small SEM. In this case, many more than 100 cells have good sampling I would presume, given the fairly small SEM.

These are complicated statistical issues that take time and effort to learn.

Wow way to discombobulate my statement...real statement was...So as the sample size decrease, the normal calculation of the SEM progressively underestimates the SEM more and more. I can do the math for you to show how the SEM underestimates as sample size decreases.

Anyways since you obviously need me to walk you through this in “Introduction to Engineering Experimentation” (Wheeler/Ganji, 1996) chapter 6.4 applied here:

For a given gridcell(cube) the sample size is very small with respect to spatial variance; therefore for each cell we must use the Student’s t distribution functions (dependent on the number of samples and confidence required) to estimate a particular cell’s confidence interval at the time of measurement. This states that the ½ confidence interval equals the t-distribution value times the sampled standard deviation divided by the square root of the number of samples.

Example...let’s say a cell has three concurrent samples values of 285.00(Kelvin), 286.00(Kelvin) and 287.00(Kelvin). The mean is (285.00+286.00+287.00)/3= 286.00K. The sampled (though not necessarily the true) standard deviation is (((285-286)^2+(286-286)^2+(287-286)^2)/2)^0.5 = 1.00K. 95% confidence corresponds to a probability of the mean temperature falling outside of the confidence interval of 0.05. Using a Student’s t- table, degrees of freedom = #samples -1 = 3-1=2. The t-distribution from the table @ 2 degrees of freedom and .05/2= .025 is 4.303. The estimated ½ confidence interval is the t distribution times the sampled standard deviation divided by the square root of the number of samples. This is 4.303*1.00K/(3)^0.5 = 2.48K. In other words, with only this data in hand, the best estimate @ 95% confidence is 286K +/- 2.48K. Likewise @ 90% confidence the result would be 286K +/- 1.69K. So even if you had hundredths of degree accuracy on the measuring device, your precision is wiped out by the spatial variance combined with low sample size. If your other cells have similar sample sizes and standard deviations, further processing to get the global average won’t improve the approximate size of this error.

Now let’s work this backwards; we want to estimate how many sample sizes we need to get +/-0.1K @95% confidence. We don’t know the number of samples yet so we have to estimate using an initial guess and our low sample size data. I’ll assume the sample size is large (>30) so we get to use the normal distribution, using 1K as an estimate for standard deviation we get an estimate of (1.96*1K/.1K)^2 = 384 samples (in that grid cell at that time). So if you have a standard deviation of 1.00K and you want to achieve +/-0.1K @95% confidence, based on the three samples we took we estimate that we will need 384 samples to achieve that.

So out of 2500 cells only 5% have 3 or more observations...so either way you need a lot more than 3 obs to justify the measurement error they claim.

Please go back and relearn your statistics.

[skierinvermont]

The appropriate method is more complicated than you have presented and the example you have given is inapplicable to the situation at hand. We are not trying to establish climatologies for cubes of water. If there are repeated measurements from the same depth and location, the measurements themselves have zero SEM (except for the precision of the instruments themselves).

In order to estimate the change in OHC globally I do not need to have high confidence estimates of the actual temperature in every 1 X 1 degree grid cell (which is the example you have given). All I need is about 100 locations worldwide where I have have measurements about once or twice a decade at various depth levels (maybe 50, 100, 200, 400, 700, 1000, 1500, 2000). This would leave the vast majority (about 99%) of 1X1 degree grid cells completely unsampled.

it's just like for surface temperature. The vast majority of 1X1 degree grid cells are not sampled at all. Zero observations.

I suggest you read the statistical methods used to calculate the SEM in the paper and point out how you find them deficient.